# Quantum mechanics - spin operator eigenvalue probability?

Hi,
I have this problem on a past exam paper I am having some trouble with:

"in the conventional basis of the eigenstates of the Sz operator, the spin state of a spin-1/2 particle is described by the vector:

$$u = \left( \stackrel{cos a}{e^i^b sina} \right)$$ where a and B are constants.

find the probability that a measurement of the y-component of the spin of the particle will yield the result $$0.5\hbar$$ ."

For the life of me I cannot work out how to write out matrices legibly on this thing, so I will summarize what is bothering me. I am given pauli matrices $$\sigma_x_,_y_,_z$$that I cannot write out properly, and the spin operator is given by $$S_i = i\hbar\sigma_i$$.

In the question I am given the vector u, which is apparently expressed in the basis of Sz eigenstates.

Am I justified in putting this vector u into an eigenvalue equation $$S_{y} u = a_{y} u$$ ,

where ay is my eigenvalue, when the vector I would be operating on is made from a basis of eigenstates of another operator (Sz)?

I tried this and got two equations for ay, neither of which gives $$a_y = 0.5\hbar$$.

does this mean I can conclude that there is zero probability of finding the y-component of the spin being equal to $$0.5\hbar$$ ?

or do I somehow have to wangle it so that I get another vector (not u) that is in the Sy eigenstate basis?

thanks.

PS. sorry, this crazy thing will not let me change something 5 lines up where I should have said
"Am I justified in putting this vector u into an eigenvalue equation S_y u = 0.5\hbar u"

Last edited:

gabbagabbahey
Homework Helper
Gold Member
"Am I justified in putting this vector u into an eigenvalue equation S_y u = 0.5\hbar u"
You tell me. If $u$ is an eigenvector of $S_z$, is it also an eigenvector of S_y[/itex]?

You tell me. If $u$ is an eigenvector of $S_z$, is it also an eigenvector of S_y[/itex]?
are you actually asking me, do these two operators commute? and if their commutator [Sy, Sz] = 0, then yes they do?

is that right, if the eigenvector of one operator is equal to the eigenvector of another, then the operators commute?

gabbagabbahey
Homework Helper
Gold Member
if the eigenvector of one operator is equal to the eigenvector of another, then the operators commute?
Yes. So, do $S_z$ and $S_y$ commute?

Yes. So, do $S_z$ and $S_y$ commute?
not by my calculations, although I just woke up 5 minutes ago. I get [Sy,Sz] is equal to
(hbar^2)/4 multiplied by a 2x2 matrix containing zeros on the leading diagonal and 2i elsewhere.

gabbagabbahey
Homework Helper
Gold Member
Right, they don't commute; $[S_y,S_z]=i\hbar S_x$. So, $u$ is clearly not an eigenstate of $S_y$.

What are the eigenstates of $S_y$?

Right, they don't commute; $[S_y,S_z]=i\hbar S_x$. So, $u$ is clearly not an eigenstate of $S_y$.

What are the eigenstates of $S_y$?
i got that one of them was the column vector, say, x:
x_1

ix_1

and the other vector y:
y_1

-iy_1

but I was not convinced by this because, if the upper component is say, x1, and I choose x1 = 1, then if I go to normalize it i do
1/sqrt[(1^2) + (i^2)] = 1/(1-1) = 0, so how can this be right if Im getting a normalization constant of zero?

vela
Staff Emeritus
You forgot to take the conjugate. You should be calculating $\langle x|x \rangle = x^\dagger x$, not $x^T x$.